# Irrational Equation A5

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https://algebrica.org/exercises/irrational-equation-a-5/
Fetched from algebrica.org test 14699; source modified 2025-04-25T17:22:46.

Before you begin, review the key concepts behind [irrational equations](../irrational-equations.md) to better follow the solution process.

Solve the irrational equation:

\\[
\\sqrt\[3\]{2x^2+x-3} = \\sqrt\[3\]{3}
\\]

* * *

Determine the set of admissible solutions. The equation is in the form:

\\[
\\sqrt\[n\]{f(x)} = k
\\]

 where \\(n\\), the index of the root is odd and \\(k\\) is a non-negative number. Since the index of the root is an odd number, no existence conditions need to be imposed because the root exists within the set of real numbers.

Raise both members to the cube:

\\[
\\begin{align} &2x^2 +x -3 = 3 \\\\\[1ex\] &2x^2 +x -3-3 = 0\\\\\[1ex\] &2x^2 +x -6 = 0\\\\\[1ex\] \\end{align}
\\]

* * *

We have obtained a [second-degree](../quadratic-equations.md) equation that can be solved using the [quadratic formula](../quadratic-formula.md):

\\[
\\begin{align\*} x\_{1,2} &= \\frac{-1 \\pm \\sqrt{1^2 -4(2)(-6)}}{2(2)} \\\\\[1ex\] x\_{1,2} &= \\frac{-1 \\pm \\sqrt{49}}{4}\\\\\[1ex\] x\_{1,2} &= \\frac{-1 \\pm 7}{4}\\\\\[1ex\] \\end{align\*}
\\]

 We have:

\\[
x = -\\frac{8}{4} \\to x = -2
\\]



\\[
x = \\frac{6}{4} \\to x = \\frac{3}{2}
\\]

 The solution to the equation is:

\\[
x = -2 \\quad x= \\frac{3}{2}
\\]

* * *

Check if the solution satisfy the initial equation by substituting the value in the original equation. Remember that this step is crucial because solutions may occur in the set of acceptable solutions but do not satisfy the identity of the equation.

For \\(x = -2 \\) we have:

\\[
\\begin{align} &\\sqrt\[3\]{2(-2)^2 +(-2) -3} = \\sqrt\[3\]{3}\\\\\[1ex\] &\\sqrt\[3\]{8 -2 -3} = \\sqrt\[3\]{3}\\\\\[1ex\] &\\sqrt\[3\]{3} = \\sqrt\[3\]{3} \\end{align}
\\]

 The equality is verified, so \\(x\\) is a solution to the equation.

For \\(x = 3/2 \\) we have:

\\[
\\begin{align} &\\sqrt\[3\]{2(3/2)^2 +(3/2) -3} = \\sqrt\[3\]{3}\\\\\[1ex\] &\\sqrt\[3\]{2 \\cdot 9/4 +3/2 -3} = \\sqrt\[3\]{3}\\\\\[1ex\] &\\sqrt\[3\]{9/2 +3/2 -3} = \\sqrt\[3\]{3}\\\\\[1ex\] &\\sqrt\[3\]{12/2 -3} = \\sqrt\[3\]{3}\\\\\[1ex\] &\\sqrt\[3\]{6-3} = \\sqrt\[3\]{3}\\\\\[1ex\] &\\sqrt\[3\]{3} = \\sqrt\[3\]{3}\\\\\[1ex\] \\end{align}
\\]

 The equality is verified, so \\(x\\) is a solution to the equation.

The solution to the equation is:

\\[
x= -2 \\quad \\quad x = \\frac{3}{2}
\\]